Question

If $y=x^{2}f(x),8f(x)+6f\left(\frac{1}{x}\right)=x+5$ and ${\left(\frac{dy}{dx}\right)}_{x=-1}=\frac{1}{a}$, then find the value of $a$.

Answer 1

Answer: -14

$8f(x)+6f\left(\frac{1}{x}\right)=x+5$ ...(i)
Replacing $x$ by $\frac{1}{x}$, we get
$8f\left(\frac{1}{x}\right)+6f(x)=\frac{1}{x}+5$ ...(ii)
Solving (i.) and (ii), we get
$f(x)=\frac{1}{14}\left(4x-\frac{3}{x}+5\right)$
$y=x^{2}f(x)$
$=x^2\left[\frac{1}{14}\left(4x-\frac{3}{x}+5\right)\right]$
$\Rightarrow y=\frac{1}{14}\left(4x^3-3x+5x^2\right)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{14}\left(12x^2-3+10x\right)$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=-1}=\frac{1}{14}(12-3-10)=\frac{-1}{14}$
$\Rightarrow a=-14$